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1 year ago

Find the Diameter of the Circle with the following equation. Round to the nearest tenth.

Mathematics

2 answers:

CaHeK987 [17]1 year ago

8 0

Answer:

11.8 units

Step-by-step explanation:

The circle equation is given as:

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:(x - a)^2 + (y - b)^2 = r^2

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:(x - a)^2 + (y - b)^2 = r^2Where

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:(x - a)^2 + (y - b)^2 = r^2WhereDiameter = 2r

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:(x - a)^2 + (y - b)^2 = r^2WhereDiameter = 2rBy comparing both equations, we have

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:(x - a)^2 + (y - b)^2 = r^2WhereDiameter = 2rBy comparing both equations, we haver^2 = 35

The circle equation is given as:(x - 2)^2 + (y + 6)^2 = 35A circle equation is represented as:(x - a)^2 + (y - b)^2 = r^2WhereDiameter = 2rBy comparing both equations, we haver^2 = 35Take the square root of both sides

Free_Kalibri [48]1 year ago

6 0

Answer: 8,9.

Step-by-step explanation:

$(x-2)^2+(y+6)^2=20\\Circle \ equation\\\boxed {(x-a)^2+(y-b)^2=r^2}\\r^2=20\\r*r=\sqrt{20}*\sqrt{20}\\ r=\sqrt{20}\\ r=\sqrt{4*5}\\ r=2\sqrt{5} \\ D=2r\\D=2*2\sqrt{5}\\ D=4\sqrt{5}\\ D\approx8,9.$

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X/5 - 12 = -8 equals what

Luda [366]

The answer is X = 20

5 0

3 years ago

Read 2 more answers

How many zeroes do we write when we write all the integers 1 to 243 in base 3?

Monica [59]

Answer:

289 numbers

Step-by-step explanation:

Above you will find the list of integers from 1 to 243 in base 3:

(1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002, 1010, 1011, 1012, 1020, 1021, 1022, 1100, 1101, 1102, 1110, 1111, 1112, 1120, 1121, 1122, 1200, 1201, 1202, 1210, 1211, 1212, 1220, 1221, 1222, 2000, 2001, 2002, 2010, 2011, 2012, 2020, 2021, 2022, 2100, 2101, 2102, 2110, 2111, 2112, 2120, 2121, 2122, 2200, 2201, 2202, 2210, 2211, 2212, 2220, 2221, 2222, 10000, 10001, 10002, 10010, 10011, 10012, 10020, 10021, 10022, 10100, 10101, 10102, 10110, 10111, 10112, 10120, 10121, 10122, 10200, 10201, 10202, 10210, 10211, 10212, 10220, 10221, 10222, 11000, 11001, 11002, 11010, 11011, 11012, 11020, 11021, 11022, 11100, 11101, 11102, 11110, 11111, 11112, 11120, 11121, 11122, 11200, 11201, 11202, 11210, 11211, 11212, 11220, 11221, 11222, 12000, 12001, 12002, 12010, 12011, 12012, 12020, 12021, 12022, 12100, 12101, 12102, 12110, 12111, 12112, 12120, 12121, 12122, 12200, 12201, 12202, 12210, 12211, 12212, 12220, 12221, 12222, 20000, 20001, 20002, 20010, 20011, 20012, 20020, 20021, 20022, 20100, 20101, 20102, 20110, 20111, 20112, 20120, 20121, 20122, 20200, 20201, 20202, 20210, 20211, 20212, 20220, 20221, 20222, 21000, 21001, 21002, 21010, 21011, 21012, 21020, 21021, 21022, 21100, 21101, 21102, 21110, 21111, 21112, 21120, 21121, 21122, 21200, 21201, 21202, 21210, 21211, 21212, 21220, 21221, 21222, 22000, 22001, 22002, 22010, 22011, 22012, 22020, 22021, 22022, 22100, 22101, 22102, 22110, 22111, 22112, 22120, 22121, 22122, 22200, 22201, 22202, 22210, 22211, 22212, 22220, 22221, 22222, 100000)

If you count them, you will find that there are 289 numbers in total!

8 0

3 years ago

Please help me find all the missing angles in the two problems

Yakvenalex [24]

the answer for the second problem is x=99 degrees and y= 261 degrees

dunno the first problem :/

3 0

3 years ago

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9

mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$

$16\cdot t^{2}-90\cdot t +81 = 0$ (1)

By Quadratic Formula:

$t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}$

$t_{1} = 4.5\,s$ , $t_{2} = 1.125\,s$

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0

3 years ago

O is the centre of the circle, EF is a tangent, angle BCE = 28°, angle ACD = 31°

andriy [413]

Answer

a. 28˚

b. 76˚

c. 104˚

d. 56˚

Step-by-step explanation

Given,

∠BCE=28° ∠ACD=31° & line AB=AC .

According To the Question,

a. the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.(Alternate Segment Theorem) Thus, ∠BAC=28°

b. We Know The Sum Of All Angles in a triangle is 180˚, 180°-∠CAB(28°)=152° and ΔABC is an isosceles triangle, So 152°/2=76˚

thus , ∠ABC=76° .

c. We know the Sum of all angles in a triangle is 180° and opposite angles in a cyclic quadrilateral(ABCD) add up to 180˚,

Thus, ∠ACD + ∠ACB = 31° + 76° ⇔ 107°

Now, ∠DCB + ∠DAB = 180°(Cyclic Quadrilateral opposite angle)

∠DAB = 180° - 107° ⇔ 73°

& We Know, ∠DAC+∠CAB=∠DAB ⇔ ∠DAC = 73° - 28° ⇔ 45°

Now, In Triangle ADC Sum of angles in a triangle is 180°

∠ADC = 180° - (31° + 45°) ⇔ 104˚

d. ∠COB = 28°×2 ⇔ 56˚ , because With the Same Arc(CB) The Angle at circumference are half of the angle at the centre

For Diagram, Please Find in Attachment

4 0

3 years ago